Ash*_*shh 4 mongoose mongodb node.js mongodb-query aggregation-framework
我有以下收藏
[
{
"setting": "Volume",
"_id": ObjectId("5a934e000102030405000000"),
"counting": 1
},
{
"setting": "Brightness",
"_id": ObjectId("5a934e000102030405000001"),
"counting": 1
},
{
"setting": "Contrast",
"_id": ObjectId("5a934e000102030405000002"),
"counting": 1
},
{
"setting": "Contrast",
"_id": ObjectId("5a934e000102030405000003"),
"counting": 1
},
{
"setting": "Contrast",
"_id": ObjectId("5a934e000102030405000004"),
"counting": 0
},
{
"setting": "Sharpness",
"_id": ObjectId("5a934e000102030405000005"),
"counting": 1
},
{
"setting": "Sharpness",
"_id": ObjectId("5a934e000102030405000006"),
"counting": 1
},
{
"setting": "Language",
"_id": ObjectId("5a934e000102030405000007"),
"counting": 1
},
{
"setting": "Language",
"_id": ObjectId("5a934e000102030405000008"),
"counting": 0
}
]
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现在我想group通过setting并且只想要结果中最靠前的两个数据useless
所以我的输出应该是sort通过计数
[
{
"setting": "Contrast",
"counting": 2
},
{
"setting": "Sharpness",
"counting": 2
},
{
"setting": "Useless",
"counting": 3
}
]
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如果你可以侥幸逃脱,那么最好将简化的结果“填充”到一个文档中,然后$slice是前两个和$sum其余的:
Model.aggregate([
{ "$group": {
"_id": "$setting",
"counting": { "$sum": "$counting" }
}},
{ "$sort": { "counting": -1 } },
{ "$group": {
"_id": null,
"data": { "$push": "$$ROOT" }
}},
{ "$addFields": {
"data": {
"$let": {
"vars": { "top": { "$slice": ["$data", 0, 2 ] } },
"in": {
"$concatArrays": [
"$$top",
{ "$cond": {
"if": { "$gt": [{ "$size": "$data" }, 2] },
"then":
[{
"_id": "Useless",
"counting": {
"$sum": {
"$map": {
"input": {
"$filter": {
"input": "$data",
"cond": { "$not": { "$in": [ "$$this._id", "$$top._id" ] } }
}
},
"in": "$$this.counting"
}
}
}
}],
"else": []
}}
]
}
}
}
}},
{ "$unwind": "$data" },
{ "$replaceRoot": { "newRoot": "$data" } }
])
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如果它可能是一个非常“大”的结果甚至减少了,那么对“其余”$limit使用 a $facet:
Model.aggregate([
{ "$facet": {
"top": [
{ "$group": {
"_id": "$setting",
"counting": { "$sum": "$counting" }
}},
{ "$sort": { "counting": -1 } },
{ "$limit": 2 }
],
"rest": [
{ "$group": {
"_id": "$setting",
"counting": { "$sum": "$counting" }
}},
{ "$sort": { "counting": -1 } },
{ "$skip": 2 },
{ "$group": {
"_id": "Useless",
"counting": { "$sum": "$counting" }
}}
]
}},
{ "$project": {
"data": {
"$concatArrays": [
"$top","$rest"
]
}
}},
{ "$unwind": "$data" },
{ "$replaceRoot": { "newRoot": "$data" } }
])
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甚至$lookup使用 MongoDB 3.6:
Model.aggregate([
{ "$group": {
"_id": "$setting",
"counting": { "$sum": "$counting" }
}},
{ "$sort": { "counting": -1 } },
{ "$limit": 2 },
{ "$group": {
"_id": null,
"top": { "$push": "$$ROOT" }
}},
{ "$lookup": {
"from": "colllection",
"let": { "settings": "$top._id" },
"pipeline": [
{ "$match": {
"$expr": {
"$not": { "$in": [ "$setting", "$$settings" ] }
}
}},
{ "$group": {
"_id": "Useless",
"counting": { "$sum": "$counting" }
}}
],
"as": "rest"
}},
{ "$project": {
"data": {
"$concatArrays": [ "$top", "$rest" ]
}
}},
{ "$unwind": "$data" },
{ "$replaceRoot": { "newRoot": "$data" } }
])
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几乎完全相同,并且都返回相同的结果:
{ "_id" : "Contrast", "counting" : 2 }
{ "_id" : "Sharpness", "counting" : 2 }
{ "_id" : "Useless", "counting" : 3 }
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可选地$project在每个末尾而不是$replaceRootif 对字段名称的控制对您来说非常重要。通常我只是坚持使用$group默认值
如果您的 MongoDB 早于 3.4 并且结果"Useless"余数实际上太大而无法使用第一种方法的任何变体,那么简单的Promise解决方案基本上就是答案,一个用于aggregate基本计数,另一个用于基本计数,只需进行数学计算:
let [docs, count] = await Promise.all([
Model.aggregate([
{ "$group": {
"_id": "$setting",
"counting": { "$sum": "$counting" }
}},
{ "$sort": { "counting": -1 } },
{ "$limit": 2 },
]),
Model.count().exec()
]);
docs = [
...docs,
{
"_id": "Useless",
"counting": count - docs.reduce((o,e) => o + e.counting, 0)
}
];
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或者没有async/await:
Promise.all([
Model.aggregate([
{ "$group": {
"_id": "$setting",
"counting": { "$sum": "$counting" }
}},
{ "$sort": { "counting": -1 } },
{ "$limit": 2 },
]),
Model.count().exec()
]).then(([docs, count]) => ([
...docs,
{
"_id": "Useless",
"counting": count - docs.reduce((o,e) => o + e.counting, 0)
}
]).then( result => /* do something */ )
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通过简单地运行单独的查询来计算集合项,这基本上是对古老的“总页数”方法的一种变体。
运行单独的请求通常是执行此操作的古老方式,并且通常效果最佳。其余的解决方案主要针对“聚合技巧”,因为这是您所要求的,这就是您通过显示同一事物的不同变化而得到的答案。
一种变体将所有结果放入单个文档中(在可能的情况下,当然是由于 BSON 限制),而其他变体通过以不同的形式再次运行查询而在“古老”方法上有所不同。$facet并联和$lookup串联。